Let \(p\)  be a rational prime and \(p\mathcal {O}_{K_{4}}=\mathfrak {p}_1...\mathfrak {p}_g\)  its factorization in \(\mathcal {O}_{K_{4}}\) . Denote by \(f\)  the residue degree of \(p\)  in \(K_4\) ,
and by \(k\)  the positive integer less than 16, such that \(p\equiv k \pmod {16}\) . Then, by the theorem of the cyclotomic reciprocity law (see [1]}), we have:
\({\begin{tabular}{|c|c|c|c|c|c|c|c|c|}\hline k & 1 & 3 & 5 & 7 & 9 & 11 & 13&15 \\\hline f& 1&4 & 4 & 2 & 2 &4 & 4 &2 \\\hline g& 8&2 & 2 & 4 & 4 &2 & 2 &4 \\\hline \end{tabular} }\) 
It follows that the rational primes that decompose into the product of two prime ideals of \(K_4\)  are exactly those which are congruent to 3 or \(5\pmod {8}\) . So a prime \(p\)  decomposes into the product of two prime ideals of \(K_m\) , is congruent to 3 or \(5\pmod {8}\) .
For the converse, assume that \(p\equiv 3\)  or \(5\pmod {8}\)  and \(p\mathcal {O}_{K_m}=\mathfrak {p}_1 \mathfrak {p}_2\) , for \(m\ge 4\) . As \(K_{m+1}=K_m(\sqrt{\zeta _{2^{m}}})\) , then
\(\left( \frac{\zeta _{2^{m}}}{\mathfrak {p}_i}\right)=\left( \frac{\zeta _{2^{m-1}}}{\mathfrak {p}_i}\right)=-1\) . So the result comes by induction.
